2008 Indonesia MO Problems/Problem 2

Problem

Prove that for every positive reals $x$ and $y$ , $\frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac {2}{x + y + 2}.$

Solution

By the Cauchy-Schwarz Inequality, $(1+1)(1+x) \ge (1 + \sqrt{x})^2$ and $(1+1)(1+y) \ge (1 + \sqrt{y})^2$ , with equality happening in the earlier inequality when $x = 1$ and equality happening in the latter inequality when $y = 1$ . Because $x,y > 0$ , $\begin{align*} \frac{1}{(1 + \sqrt{x})^2} &\ge \frac{1}{2+2x} \\ \frac{1}{(1 + \sqrt{y})^2} &\ge \frac{1}{2+2y} \\ \frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} &\ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}). \end{align*}$ By the AM-GM Inequality, we know that $\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}$ . For the equality case, $\frac{1}{1+x} = \frac{1}{1+y}$ , so $x = y$ . Additionally, by the AM-GM Inequality, $\frac12 \cdot (\frac{x+1}{2} + \frac{y+1}{2}) \ge \sqrt{\frac{(x+1)(y+1)}{4}}$ . For the equality case, $\frac{x+1}{2} = \frac{y+1}{2}$ , so $x = y$ . Because $x,y \ge 0$ , $\begin{align*} \frac12 \cdot (\frac{x+y+2}{2}) &\ge \frac{\sqrt{(x+1)(y+1)}}{2} \\ \frac{x+y+2}{2} &\ge \sqrt{(x+1)(y+1)} \\ \sqrt{\frac{1}{(x+1)(y+1)}} &\ge \frac{2}{x+y+2}. \end{align*}$ Therefore, since $\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y})$ and $\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}$ and $\sqrt{\frac{1}{(x+1)(y+1)}} \ge \frac{2}{x+y+2}$ , we must have $\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac{2}{x+y+2}$ , with equality happening when $x = y = 1$ .

2008 Indonesia MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 3
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2008 Indonesia MO Problems/Problem 2

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